Finding the Value of n in Binomial Expansion Middle Term

The problem asks us to find the value of \(n\) given the middle term of the binomial expansion of \({\left( {{x^2} + \frac{1}{x}} \right)^{2n}}\). The middle term is given as \(184756x^{10}\).

Understanding the Middle Term in Binomial Expansion

For a binomial expansion of the form \((a+b)^N\), the total number of terms is \(N+1\).

In our case, the expression is \({\left( {{x^2} + \frac{1}{x}} \right)^{2n}}\). Here, \(a = x^2\), \(b = \frac{1}{x}\), and \(N = 2n\). The total number of terms in this expansion is \(2n+1\). Since \(2n\) is always an even number (for any integer \(n\)), \(2n+1\) is always an odd number. An odd number of terms means there is a unique middle term.

The position of the middle term is given by \(\frac{N+1}{2} + 1\) if N is odd, or \(\frac{N}{2} + 1\) if N is even. Since N=2n is even, the middle term is at position:

Middle term position \( = \frac{2n}{2} + 1 = n + 1\)

So, the \((n+1)^{th}\) term is the middle term of the binomial expansion.

Using the General Term Formula for Binomial Expansion

The general term, \(T_{r+1}\), in the expansion of \((a+b)^N\) is given by the formula:

\(T_{r+1} = \binom{N}{r} a^{N-r} b^r\)

For the middle term, which is the \((n+1)^{th}\) term, we have \(r+1 = n+1\), so \(r=n\). Substituting \(N=2n\), \(a=x^2\), and \(b=\frac{1}{x}\) into the general term formula:

\(T_{n+1} = \binom{2n}{n} (x^2)^{2n-n} \left(\frac{1}{x}\right)^n\)

Simplifying the Middle Term

Let's simplify the expression for the middle term:

\(T_{n+1} = \binom{2n}{n} (x^2)^n (x^{-1})^n\)

Using the exponent rules \((a^m)^p = a^{mp}\) and \(a^m a^p = a^{m+p}\):

\(T_{n+1} = \binom{2n}{n} x^{2 \times n} x^{-1 \times n}\)

\(T_{n+1} = \binom{2n}{n} x^{2n} x^{-n}\)

\(T_{n+1} = \binom{2n}{n} x^{2n - n}\)

\(T_{n+1} = \binom{2n}{n} x^{n}\)

Comparing with the Given Middle Term Value

We are given that the middle term is \(184756x^{10}\). We have calculated the middle term to be \(\binom{2n}{n} x^{n}\). By comparing the two expressions for the middle term, we can equate the coefficients and the powers of \(x\).

\(\binom{2n}{n} x^{n} = 184756x^{10}\)

Equating Powers of x

The power of \(x\) on the left side is \(n\), and the power of \(x\) on the right side is \(10\). Equating the powers:

\(n = 10\)

Equating Coefficients

The coefficient on the left side is \(\binom{2n}{n}\), and the coefficient on the right side is \(184756\). Equating the coefficients:

\(\binom{2n}{n} = 184756\)

Now, let's substitute the value of \(n=10\) that we found:

\(\binom{2 \times 10}{10} = \binom{20}{10}\)

Let's calculate the value of \(\binom{20}{10}\) to verify if it equals \(184756\).

\(\binom{20}{10} = \frac{20!}{10! (20-10)!} = \frac{20!}{10! 10!}\)

\(\binom{20}{10} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\)

We can simplify this calculation:

• \(10 \times 2 = 20\), cancels with 20.
• \(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 1 = 362880\)
• Numerator \(20 \times 19 \times \dots \times 11\)
• Denominator \(10!\)

Performing the calculation:

\(\binom{20}{10} = 19 \times \frac{18}{9 \times 2} \times 17 \times \frac{16}{8 \times 4 \times 2} \times \frac{15}{5 \times 3} \times \frac{14}{7} \times 13 \times \frac{12}{6} \times 11\)

\(\binom{20}{10} = 19 \times 1 \times 17 \times \frac{16}{64} \times 1 \times 2 \times 13 \times 2 \times 11\)

A better simplification approach:

\(\binom{20}{10} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\)

• \(10 \times 2\) cancels \(20\)
• \(9\) cancels \(18\) (leaving 2)
• \(8\) cancels \(16\) (leaving 2)
• \(7\) cancels \(14\) (leaving 2)
• \(6\) cancels \(12\) (leaving 2)
• \(5 \times 3\) cancels \(15\) (leaving 1)
• \(4\) cancels \(2 \times 2\) (from 16 and 12)
• Remaining denominator is \(1\).

So, \(\binom{20}{10} = 19 \times 2 \times 17 \times \frac{16}{8 \times 4 \times 2} \times 1 \times \frac{14}{7} \times 13 \times \frac{12}{6} \times 11 / (10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)\) -- Let's restart the calculation simplification carefully.

Denominator = \(10! = 3628800\)

Numerator = \(20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 = 670442572800\)

\(\binom{20}{10} = \frac{670442572800}{3628800} = 184756\)

The calculated value of \(\binom{20}{10}\) is indeed \(184756\). This confirms that our value of \(n=10\) is correct, as it satisfies both the power of \(x\) and the coefficient of the middle term.

Conclusion: Value of n

Based on the analysis of the middle term of the binomial expansion \({\left( {{x^2} + \frac{1}{x}} \right)^{2n}}\) and comparing it with the given value \(184756x^{10}\), we determined that the power of \(x\) in the middle term is \(n\) and the coefficient is \(\binom{2n}{n}\). By equating the powers of \(x\), we found \(n=10\). This value was confirmed by calculating \(\binom{20}{10}\) which matched the given coefficient.

Thus, the value of \(n\) is \(10\).

Binomial Expansion Revision Table

Additional Information on Binomial Terms

Understanding the terms in a binomial expansion is crucial for solving problems like finding a specific term, the middle term, or the term independent of \(x\). Let's explore a few related points:

• Term Independent of x: This is the term where the power of \(x\) (or the variable) becomes zero. You find \(r\) such that the variable part \(x^{N-r} (x^{-1})^r\) simplifies to \(x^0\).
• Coefficient Calculation: Binomial coefficients \(\binom{N}{r}\) can be found using Pascal's triangle or the factorial formula. For large \(N\), the factorial formula is used, as seen in the solution.
• Sum of Coefficients: The sum of all binomial coefficients in the expansion of \((a+b)^N\) is \(2^N\). If the expression is \((1+x)^N\), the sum is found by setting \(x=1\), giving \(2^N\).
• Properties of Coefficients: Coefficients are symmetric, i.e., \(\binom{N}{r} = \binom{N}{N-r}\). The largest coefficient occurs at the middle term(s).

The problem required identifying the middle term position and then applying the general term formula, equating the result to the given information to solve for the unknown variable \(n\).